How would a composite variable be strongly correlated with one variable but not the other?











up vote
2
down vote

favorite












I have two variables x1 and x2 which measure relatively similar things (r ~ 0.6), with x2 slightly larger than x1 on average. I then created a new variable x3 by subtracting the two: x3 = x1 - x2.



However, when I ran the Pearson correlations, x3 is strongly negatively correlated with x2 as expected (r ~ -0.6), but x3 is not very correlated with x1 (r ~ 0.1). How is this possible?










share|cite|improve this question




















  • 2




    A scatter plot matrix should help.
    – Nick Cox
    14 hours ago






  • 1




    Possible duplicate of When A and B are positively related variables, can they have opposite effect on their outcome variable C?
    – sds
    14 hours ago















up vote
2
down vote

favorite












I have two variables x1 and x2 which measure relatively similar things (r ~ 0.6), with x2 slightly larger than x1 on average. I then created a new variable x3 by subtracting the two: x3 = x1 - x2.



However, when I ran the Pearson correlations, x3 is strongly negatively correlated with x2 as expected (r ~ -0.6), but x3 is not very correlated with x1 (r ~ 0.1). How is this possible?










share|cite|improve this question




















  • 2




    A scatter plot matrix should help.
    – Nick Cox
    14 hours ago






  • 1




    Possible duplicate of When A and B are positively related variables, can they have opposite effect on their outcome variable C?
    – sds
    14 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have two variables x1 and x2 which measure relatively similar things (r ~ 0.6), with x2 slightly larger than x1 on average. I then created a new variable x3 by subtracting the two: x3 = x1 - x2.



However, when I ran the Pearson correlations, x3 is strongly negatively correlated with x2 as expected (r ~ -0.6), but x3 is not very correlated with x1 (r ~ 0.1). How is this possible?










share|cite|improve this question















I have two variables x1 and x2 which measure relatively similar things (r ~ 0.6), with x2 slightly larger than x1 on average. I then created a new variable x3 by subtracting the two: x3 = x1 - x2.



However, when I ran the Pearson correlations, x3 is strongly negatively correlated with x2 as expected (r ~ -0.6), but x3 is not very correlated with x1 (r ~ 0.1). How is this possible?







correlation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago









Nick Cox

37.9k480127




37.9k480127










asked 20 hours ago









hlinee

387




387








  • 2




    A scatter plot matrix should help.
    – Nick Cox
    14 hours ago






  • 1




    Possible duplicate of When A and B are positively related variables, can they have opposite effect on their outcome variable C?
    – sds
    14 hours ago














  • 2




    A scatter plot matrix should help.
    – Nick Cox
    14 hours ago






  • 1




    Possible duplicate of When A and B are positively related variables, can they have opposite effect on their outcome variable C?
    – sds
    14 hours ago








2




2




A scatter plot matrix should help.
– Nick Cox
14 hours ago




A scatter plot matrix should help.
– Nick Cox
14 hours ago




1




1




Possible duplicate of When A and B are positively related variables, can they have opposite effect on their outcome variable C?
– sds
14 hours ago




Possible duplicate of When A and B are positively related variables, can they have opposite effect on their outcome variable C?
– sds
14 hours ago










4 Answers
4






active

oldest

votes

















up vote
13
down vote













Here's a simple example. Suppose $ε_1$ and $ε_2$ are independent standard normal random variables. Define $X_1 = ε_1$, $X_2 = X_1 + ε_2$, and $X_3 = X_1 - X_2$. The correlation of $X_1$ with $X_2$ is then $tfrac{1}{sqrt{2}} approx .71$. Likewise, the correlation of $X_2$ with $X_3$ is $-tfrac{1}{sqrt{2}}$. But the correlation of $X_1$ with $X_3$ is the correlation of $ε_1$ with $ε_1 - (ε_1 + ε_2) = -ε_2$, which is 0 since the $ε_i$s are independent.






share|cite|improve this answer






























    up vote
    2
    down vote













    This is by construction of $x_3$. Given that $x_2$ and $x_1$ are closely related - in terms of their Pearson correlation if you subtract one from the other, you reduce correlation. The best way to see that is to consider the extreme scenario of complete correlation, i.e., $x_2=x_1$, in which case $x_3=x_1-x_2=0$, which is fully deterministic, i.e., $rapprox 0$.



    You can do a more formal argument using the definition of the Pearson correlation by looking at the covariation between $x_3$ and $x_1$. You will see that the covariation will be reduced. By how much, depends on the correlation between $x_1$ and $x_2$, i.e., $r_{12}$ and their standard deviations. Everything being equal, the larger $r_{12}$, the smaller $r_{13}$.






    share|cite|improve this answer








    New contributor




    Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • By "covariation", do you mean "covariance"?
      – Kodiologist
      18 hours ago


















    up vote
    1
    down vote













    You can rewrite your equation $x_3=x_2-x_1$ as $x_2=x_3-x_1$. Then regardless of what you pick as $x_1$ and $x_3$, you will have that $x_2$ is correlated to $x_1$ and $x_3$, but there is no reason to expect $x_1$ and $x_3$ to be correlated to each other. For instance, if $x_1$= number of letters in title of Best Picture Oscar winner, $x_3$= number of named hurricanes, $x_2$= number of named hurricanes - number of letters in title of Best Picture Oscar winner, then you will have that $x_3=x_2-x_1$, but that doesn't mean that $x_3$ will be correlated with $x_1$.






    share|cite|improve this answer




























      up vote
      1
      down vote













      Let $Var(X_1) = sigma_1^2$, $Var(X_2) = sigma_2^2$, and $Cov(X_1,X_2)=sigma_{12} = rhosigma_1sigma_2$
      Then $Var(X_3=X_1-X_2)=sigma_1^2+sigma_2^2 - 2sigma_{12}$



      $Cov(X_1,X_3)=sigma_1^2-sigma_{12}$



      $Cov(X_2,X_3) =sigma_{12}-sigma_2^2$



      $Corr(X_1,X_3) =frac{sigma_1^2-sigma_{12}}{sqrt{sigma_1^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



      $Corr(X_2,X_3) =frac{-sigma_2^2+sigma_{12}}{sqrt{sigma_2^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



      So $|Corr(X_1,X_3)| lt text {or} = text {or} gt |Corr(X_2,X_3)|$ depends on $sigma_1^2$ and $sigma_2^2$



      This relation cannot be determined by correlation coefficient.






      share|cite|improve this answer























        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "65"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f381477%2fhow-would-a-composite-variable-be-strongly-correlated-with-one-variable-but-not%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        13
        down vote













        Here's a simple example. Suppose $ε_1$ and $ε_2$ are independent standard normal random variables. Define $X_1 = ε_1$, $X_2 = X_1 + ε_2$, and $X_3 = X_1 - X_2$. The correlation of $X_1$ with $X_2$ is then $tfrac{1}{sqrt{2}} approx .71$. Likewise, the correlation of $X_2$ with $X_3$ is $-tfrac{1}{sqrt{2}}$. But the correlation of $X_1$ with $X_3$ is the correlation of $ε_1$ with $ε_1 - (ε_1 + ε_2) = -ε_2$, which is 0 since the $ε_i$s are independent.






        share|cite|improve this answer



























          up vote
          13
          down vote













          Here's a simple example. Suppose $ε_1$ and $ε_2$ are independent standard normal random variables. Define $X_1 = ε_1$, $X_2 = X_1 + ε_2$, and $X_3 = X_1 - X_2$. The correlation of $X_1$ with $X_2$ is then $tfrac{1}{sqrt{2}} approx .71$. Likewise, the correlation of $X_2$ with $X_3$ is $-tfrac{1}{sqrt{2}}$. But the correlation of $X_1$ with $X_3$ is the correlation of $ε_1$ with $ε_1 - (ε_1 + ε_2) = -ε_2$, which is 0 since the $ε_i$s are independent.






          share|cite|improve this answer

























            up vote
            13
            down vote










            up vote
            13
            down vote









            Here's a simple example. Suppose $ε_1$ and $ε_2$ are independent standard normal random variables. Define $X_1 = ε_1$, $X_2 = X_1 + ε_2$, and $X_3 = X_1 - X_2$. The correlation of $X_1$ with $X_2$ is then $tfrac{1}{sqrt{2}} approx .71$. Likewise, the correlation of $X_2$ with $X_3$ is $-tfrac{1}{sqrt{2}}$. But the correlation of $X_1$ with $X_3$ is the correlation of $ε_1$ with $ε_1 - (ε_1 + ε_2) = -ε_2$, which is 0 since the $ε_i$s are independent.






            share|cite|improve this answer














            Here's a simple example. Suppose $ε_1$ and $ε_2$ are independent standard normal random variables. Define $X_1 = ε_1$, $X_2 = X_1 + ε_2$, and $X_3 = X_1 - X_2$. The correlation of $X_1$ with $X_2$ is then $tfrac{1}{sqrt{2}} approx .71$. Likewise, the correlation of $X_2$ with $X_3$ is $-tfrac{1}{sqrt{2}}$. But the correlation of $X_1$ with $X_3$ is the correlation of $ε_1$ with $ε_1 - (ε_1 + ε_2) = -ε_2$, which is 0 since the $ε_i$s are independent.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 18 hours ago

























            answered 19 hours ago









            Kodiologist

            16.5k22953




            16.5k22953
























                up vote
                2
                down vote













                This is by construction of $x_3$. Given that $x_2$ and $x_1$ are closely related - in terms of their Pearson correlation if you subtract one from the other, you reduce correlation. The best way to see that is to consider the extreme scenario of complete correlation, i.e., $x_2=x_1$, in which case $x_3=x_1-x_2=0$, which is fully deterministic, i.e., $rapprox 0$.



                You can do a more formal argument using the definition of the Pearson correlation by looking at the covariation between $x_3$ and $x_1$. You will see that the covariation will be reduced. By how much, depends on the correlation between $x_1$ and $x_2$, i.e., $r_{12}$ and their standard deviations. Everything being equal, the larger $r_{12}$, the smaller $r_{13}$.






                share|cite|improve this answer








                New contributor




                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.


















                • By "covariation", do you mean "covariance"?
                  – Kodiologist
                  18 hours ago















                up vote
                2
                down vote













                This is by construction of $x_3$. Given that $x_2$ and $x_1$ are closely related - in terms of their Pearson correlation if you subtract one from the other, you reduce correlation. The best way to see that is to consider the extreme scenario of complete correlation, i.e., $x_2=x_1$, in which case $x_3=x_1-x_2=0$, which is fully deterministic, i.e., $rapprox 0$.



                You can do a more formal argument using the definition of the Pearson correlation by looking at the covariation between $x_3$ and $x_1$. You will see that the covariation will be reduced. By how much, depends on the correlation between $x_1$ and $x_2$, i.e., $r_{12}$ and their standard deviations. Everything being equal, the larger $r_{12}$, the smaller $r_{13}$.






                share|cite|improve this answer








                New contributor




                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.


















                • By "covariation", do you mean "covariance"?
                  – Kodiologist
                  18 hours ago













                up vote
                2
                down vote










                up vote
                2
                down vote









                This is by construction of $x_3$. Given that $x_2$ and $x_1$ are closely related - in terms of their Pearson correlation if you subtract one from the other, you reduce correlation. The best way to see that is to consider the extreme scenario of complete correlation, i.e., $x_2=x_1$, in which case $x_3=x_1-x_2=0$, which is fully deterministic, i.e., $rapprox 0$.



                You can do a more formal argument using the definition of the Pearson correlation by looking at the covariation between $x_3$ and $x_1$. You will see that the covariation will be reduced. By how much, depends on the correlation between $x_1$ and $x_2$, i.e., $r_{12}$ and their standard deviations. Everything being equal, the larger $r_{12}$, the smaller $r_{13}$.






                share|cite|improve this answer








                New contributor




                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                This is by construction of $x_3$. Given that $x_2$ and $x_1$ are closely related - in terms of their Pearson correlation if you subtract one from the other, you reduce correlation. The best way to see that is to consider the extreme scenario of complete correlation, i.e., $x_2=x_1$, in which case $x_3=x_1-x_2=0$, which is fully deterministic, i.e., $rapprox 0$.



                You can do a more formal argument using the definition of the Pearson correlation by looking at the covariation between $x_3$ and $x_1$. You will see that the covariation will be reduced. By how much, depends on the correlation between $x_1$ and $x_2$, i.e., $r_{12}$ and their standard deviations. Everything being equal, the larger $r_{12}$, the smaller $r_{13}$.







                share|cite|improve this answer








                New contributor




                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                share|cite|improve this answer



                share|cite|improve this answer






                New contributor




                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                answered 19 hours ago









                Gkhan Cebs

                311




                311




                New contributor




                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                New contributor





                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                Gkhan Cebs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.












                • By "covariation", do you mean "covariance"?
                  – Kodiologist
                  18 hours ago


















                • By "covariation", do you mean "covariance"?
                  – Kodiologist
                  18 hours ago
















                By "covariation", do you mean "covariance"?
                – Kodiologist
                18 hours ago




                By "covariation", do you mean "covariance"?
                – Kodiologist
                18 hours ago










                up vote
                1
                down vote













                You can rewrite your equation $x_3=x_2-x_1$ as $x_2=x_3-x_1$. Then regardless of what you pick as $x_1$ and $x_3$, you will have that $x_2$ is correlated to $x_1$ and $x_3$, but there is no reason to expect $x_1$ and $x_3$ to be correlated to each other. For instance, if $x_1$= number of letters in title of Best Picture Oscar winner, $x_3$= number of named hurricanes, $x_2$= number of named hurricanes - number of letters in title of Best Picture Oscar winner, then you will have that $x_3=x_2-x_1$, but that doesn't mean that $x_3$ will be correlated with $x_1$.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  You can rewrite your equation $x_3=x_2-x_1$ as $x_2=x_3-x_1$. Then regardless of what you pick as $x_1$ and $x_3$, you will have that $x_2$ is correlated to $x_1$ and $x_3$, but there is no reason to expect $x_1$ and $x_3$ to be correlated to each other. For instance, if $x_1$= number of letters in title of Best Picture Oscar winner, $x_3$= number of named hurricanes, $x_2$= number of named hurricanes - number of letters in title of Best Picture Oscar winner, then you will have that $x_3=x_2-x_1$, but that doesn't mean that $x_3$ will be correlated with $x_1$.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    You can rewrite your equation $x_3=x_2-x_1$ as $x_2=x_3-x_1$. Then regardless of what you pick as $x_1$ and $x_3$, you will have that $x_2$ is correlated to $x_1$ and $x_3$, but there is no reason to expect $x_1$ and $x_3$ to be correlated to each other. For instance, if $x_1$= number of letters in title of Best Picture Oscar winner, $x_3$= number of named hurricanes, $x_2$= number of named hurricanes - number of letters in title of Best Picture Oscar winner, then you will have that $x_3=x_2-x_1$, but that doesn't mean that $x_3$ will be correlated with $x_1$.






                    share|cite|improve this answer












                    You can rewrite your equation $x_3=x_2-x_1$ as $x_2=x_3-x_1$. Then regardless of what you pick as $x_1$ and $x_3$, you will have that $x_2$ is correlated to $x_1$ and $x_3$, but there is no reason to expect $x_1$ and $x_3$ to be correlated to each other. For instance, if $x_1$= number of letters in title of Best Picture Oscar winner, $x_3$= number of named hurricanes, $x_2$= number of named hurricanes - number of letters in title of Best Picture Oscar winner, then you will have that $x_3=x_2-x_1$, but that doesn't mean that $x_3$ will be correlated with $x_1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 15 hours ago









                    Acccumulation

                    1,52826




                    1,52826






















                        up vote
                        1
                        down vote













                        Let $Var(X_1) = sigma_1^2$, $Var(X_2) = sigma_2^2$, and $Cov(X_1,X_2)=sigma_{12} = rhosigma_1sigma_2$
                        Then $Var(X_3=X_1-X_2)=sigma_1^2+sigma_2^2 - 2sigma_{12}$



                        $Cov(X_1,X_3)=sigma_1^2-sigma_{12}$



                        $Cov(X_2,X_3) =sigma_{12}-sigma_2^2$



                        $Corr(X_1,X_3) =frac{sigma_1^2-sigma_{12}}{sqrt{sigma_1^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                        $Corr(X_2,X_3) =frac{-sigma_2^2+sigma_{12}}{sqrt{sigma_2^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                        So $|Corr(X_1,X_3)| lt text {or} = text {or} gt |Corr(X_2,X_3)|$ depends on $sigma_1^2$ and $sigma_2^2$



                        This relation cannot be determined by correlation coefficient.






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          Let $Var(X_1) = sigma_1^2$, $Var(X_2) = sigma_2^2$, and $Cov(X_1,X_2)=sigma_{12} = rhosigma_1sigma_2$
                          Then $Var(X_3=X_1-X_2)=sigma_1^2+sigma_2^2 - 2sigma_{12}$



                          $Cov(X_1,X_3)=sigma_1^2-sigma_{12}$



                          $Cov(X_2,X_3) =sigma_{12}-sigma_2^2$



                          $Corr(X_1,X_3) =frac{sigma_1^2-sigma_{12}}{sqrt{sigma_1^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                          $Corr(X_2,X_3) =frac{-sigma_2^2+sigma_{12}}{sqrt{sigma_2^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                          So $|Corr(X_1,X_3)| lt text {or} = text {or} gt |Corr(X_2,X_3)|$ depends on $sigma_1^2$ and $sigma_2^2$



                          This relation cannot be determined by correlation coefficient.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Let $Var(X_1) = sigma_1^2$, $Var(X_2) = sigma_2^2$, and $Cov(X_1,X_2)=sigma_{12} = rhosigma_1sigma_2$
                            Then $Var(X_3=X_1-X_2)=sigma_1^2+sigma_2^2 - 2sigma_{12}$



                            $Cov(X_1,X_3)=sigma_1^2-sigma_{12}$



                            $Cov(X_2,X_3) =sigma_{12}-sigma_2^2$



                            $Corr(X_1,X_3) =frac{sigma_1^2-sigma_{12}}{sqrt{sigma_1^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                            $Corr(X_2,X_3) =frac{-sigma_2^2+sigma_{12}}{sqrt{sigma_2^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                            So $|Corr(X_1,X_3)| lt text {or} = text {or} gt |Corr(X_2,X_3)|$ depends on $sigma_1^2$ and $sigma_2^2$



                            This relation cannot be determined by correlation coefficient.






                            share|cite|improve this answer














                            Let $Var(X_1) = sigma_1^2$, $Var(X_2) = sigma_2^2$, and $Cov(X_1,X_2)=sigma_{12} = rhosigma_1sigma_2$
                            Then $Var(X_3=X_1-X_2)=sigma_1^2+sigma_2^2 - 2sigma_{12}$



                            $Cov(X_1,X_3)=sigma_1^2-sigma_{12}$



                            $Cov(X_2,X_3) =sigma_{12}-sigma_2^2$



                            $Corr(X_1,X_3) =frac{sigma_1^2-sigma_{12}}{sqrt{sigma_1^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                            $Corr(X_2,X_3) =frac{-sigma_2^2+sigma_{12}}{sqrt{sigma_2^2(sigma_1^2+sigma_2^2 - 2sigma_{12})}}$



                            So $|Corr(X_1,X_3)| lt text {or} = text {or} gt |Corr(X_2,X_3)|$ depends on $sigma_1^2$ and $sigma_2^2$



                            This relation cannot be determined by correlation coefficient.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 9 hours ago

























                            answered 11 hours ago









                            user158565

                            4,9501317




                            4,9501317






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Cross Validated!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f381477%2fhow-would-a-composite-variable-be-strongly-correlated-with-one-variable-but-not%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Ellipse (mathématiques)

                                Quarter-circle Tiles

                                Mont Emei